Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $y = \dfrac{t^2 + 4t - 21}{t^2 - 3t} \div \dfrac{t + 7}{-8t + 32} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{t^2 + 4t - 21}{t^2 - 3t} \times \dfrac{-8t + 32}{t + 7} $ First factor the quadratic. $y = \dfrac{(t + 7)(t - 3)}{t^2 - 3t} \times \dfrac{-8t + 32}{t + 7} $ Then factor out any other terms. $y = \dfrac{(t + 7)(t - 3)}{t(t - 3)} \times \dfrac{-8(t - 4)}{t + 7} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ (t + 7)(t - 3) \times -8(t - 4) } { t(t - 3) \times (t + 7) } $ $y = \dfrac{ -8(t + 7)(t - 3)(t - 4)}{ t(t - 3)(t + 7)} $ Notice that $(t - 3)$ and $(t + 7)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ -8\cancel{(t + 7)}(t - 3)(t - 4)}{ t(t - 3)\cancel{(t + 7)}} $ We are dividing by $t + 7$ , so $t + 7 \neq 0$ Therefore, $t \neq -7$ $y = \dfrac{ -8\cancel{(t + 7)}\cancel{(t - 3)}(t - 4)}{ t\cancel{(t - 3)}\cancel{(t + 7)}} $ We are dividing by $t - 3$ , so $t - 3 \neq 0$ Therefore, $t \neq 3$ $y = \dfrac{-8(t - 4)}{t} ; \space t \neq -7 ; \space t \neq 3 $